Question

Using remainder theorem, show that the polynomial P(x) = x^3 – 2x^2 + 3x – 18 is a
multiple of x -3.

Answer 1

Answer:

$p(x)={x}^3-{2x}^2+3x-18$ is not a multiple. of $x-3$.

Step-by-step explanation:

If the question asks us to show that $p(x)={x}^3-{2x}^2+3x-18$ is the multiple of $x-3$, that in another words is proving that $x-3$ is a factor of the the polynomial $p(x)=....$. So let's use Factor Theorem to solve.

Let's find the zero of $x-3$

$x-3=0$

$x=0+3$

$x=3$

3 is the zero of the polynomial.

$p(x)={x}^3-{2x}^2+3x-18$

$p(3)={(3)}^3-{(2*3)}^2+3(3)-18$

$p(3)=27-36+9-18$

$p(3)= -18$

Answer 2

Answer:

p(x)=x 3 −2x 2 +3x−18 is not a multiple. of x-3x−3

Step-by-step explanation:

If the question asks us to show that p(x)={x}^3-{2x}^2+3x-18p(x)=x

3 −2x 2 +3x−18 is the multiple of x-3x−3 , that in another words is proving that x-3x−3 is a factor of the the polynomial p(x)=....p(x)=.... . So let's use Factor Theorem to solve.

Let's find the zero of x-3x−3

x-3=0x−3=0

x=0+3x=0+3

x=3x=3

3 is the zero of the polynomial.

p(x)={x}^3-{2x}^2+3x-18p(x)=x

3−2x 2 +3x−18

p(3)={(3)}^3-{(2*3)}^2+3(3)-18p(3)=(3)

3 −(2∗3) 2 +3(3)−18

p(3)=27-36+9-18p(3)=27−36+9−18

p(3)= -18p(3)=−18